A solid disk with a radius of #2 m# and mass of #3 kg# is rotating on a frictionless surface. If #36 W# of power is used to increase the disk's rate of rotation, what torque is applied when the disk is rotating at #8 Hz#?

1 Answer
Nov 21, 2017

Answer:

The torque is #=0.72Nm#

Explanation:

Apply the equation

#P=tau omega#

The power is #P=36W#

The frequency is #f=8Hz#

The angular velocity is

#omega=2pif=2xxpixx8=(16pi)rads^-1#

Therefore,

The torque is

#tau=P/omega=36/(16pi)=0.72Nm#